SAMPLING SAMPLING DISTRIBUTIONS DISTRIBUTIONS & CONFIDENCE & CONFIDENCE INTERVALINTERVAL

CHAPTER 3

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CONTENTCONTENT3.1 Sampling Distribution

3.2 Estimate, Estimation and Estimator

3.3 Confidence Interval for the mean μ

3.4 Confidence Interval for the Difference

between Two mean

3.5 Confidence Interval for the Proportion

3.6 Confidence Interval for the Difference

between Two Proportions

3.7 Confidence Interval for Variances and

Standard Deviations

3.8 Confidence Interval for Two Variances

and Standard Deviations

3.1 Sampling Distributions 3.1 Sampling Distributions

OBJECTIVEOBJECTIVE

After completing this chapter, you should be able to

1. Identify the sampling distribution for sample mean, difference between two sample means, sample proportion and difference between two sample proportions.

Sampling Distribution

A sampling distribution is the probability distribution, under repeated sampling of the population, of a given statistic (a numerical quantity calculated from the data values in a sample).

The formula for the sampling distribution depends on the distribution of the population, the statistic being considered, and the sample size used. A more precise formulation would speak of the distribution of the statistic for all possible samples of a given size, not just "under repeated sampling".

In other word, the sampling distribution of a statistic S for samples of size n is defined as follows: – The experiment consists of choosing a sample of size n

from the population and measuring the statistic S. The sampling distribution is the resulting probability distribution.

Sampling Distribution of Means

Imaging carrying out the following procedure:

– Take a random sample of n independent observations from a population.

– Calculate the mean of these n sample values. (Mean sample)

– Repeat the procedure until you have taken all possible samples of size n, calculating the sample mean of each one.

– Form a distribution of all the sample means.

The distribution is called sampling distribution of means

Example

Imagine that our population consists of only three numbers: the number 2, the number 3 and the number 4. Our plan is to draw an infinite number of random samples of size n = 2 and form a sampling distribution of the sample means.

Sampling Distribution for Means

The mean of the sampling distribution of means is equal to the population mean.

The standard deviation of the sampling distribution of means is

for infinite population

for finite population

If the population is normally distributed, the sampling distribution is normal regardless of sample size.

By using the Central Limit Theorem,

If the population distribution is not necessarily normal, and has mean μ and standard deviation σ , then, for sufficiently large n, the sampling distribution of is approximately normal, with mean       and standard deviation        

X

X n

2

~ ,X Nn

X

X n

1X

N n

Nn

X

EXERCISE 3.1

1. At a college, the masses of the male students can be modeled by a normal distribution with mean mass 70kg and standard deviation 5kg. Four male students are chosen at random. Find the probability that their mean mass is less than 65kg.

TIPS: Use Central Limit Theorem

Sampling Distribution for Difference Mean

2 21 2

1 1 2 21 2

2 21 2

1 2 1 21 2

2 21 2

1 2 1 21 2

For ~ , and ~ ,

~ ,

~ ,

X N X Nn n

X X Nn n

X X Nn n

n1 and n2 independent observations

EXERCISE 3.1

2. The elasticity of a polymer is affected by the concentration of a reactant. When low concentration is used, the true mean elasticity is 55, and when high concentration is used, the true mean elasticity is 60. The standard deviation of elasticity is 4, regardless of concentration.

If two random samples of sizes 16 are taken, find the probability that the difference mean of elasticity for high and low concentration is at least 2.

Sampling Distribution for Proportions

p - Proportion, Probability and Percent for population

- sample proportion of X successes in a sample of size n

- sample proportion of failures in a sample of size n

X is the binomial random variable created by counting the number of successes picked by drawing n times from the population.

The shape of the binomial distribution looks fairly Normal as long as n is large and/or p is not too extreme (not close to 0 or 1).

ˆX

pn

ˆ ˆ1q p

~ ,X bin n p

1ˆIf 1 5, then ~ ,

p pnp p p N p

n

Sampling Distribution for Proportions

The sampling distribution for proportions is a distribution of the proportions of all possible n samples that could be taken in a given situation. 

That is, the sample proportion (percent of successes in a sample), is approximately Normally distributed with

– mean p, and – standard deviation 1p p

n

EXERCISE 3.1

3. It is known that 3% of frozen pies delivered to a canteen are broken. What is the probability that, on a morning when 500 pies are delivered, 5% or more are broken?

TIPS: Use Normal Approximation to the Binomial distribution OR

Use with continuity correction

ˆ

1

p pZ

p p

n

1

2n

Sampling Distribution for Difference Proportions

1 1 2 21 1 2 2

1 2

1 1 2 21 2 1 2

1 2

1 1 2 21 2 1 2

1 2

1 1ˆ ˆIf ~ , and ~ ,

1 1ˆ ˆ ~ ,

1 1ˆ ˆ ~ ,

p p p pp N p p N p

n n

p p p pp p N p p

n n

p p p pp p N p p

n n

n1 and n2 independent observations

3.2 3.2 Estimate, Estimation, Estimator

OBJECTIVEOBJECTIVE

After completing this chapter, you should be able to

1. Define and understand the general formula of interval estimate (confidence interval) for a parameter.

Estimator

Probability function are actually families of models in the sense that each include one or more parameter.

Example: Poisson, Binomial, Normal

Any function of a random sample whose objective is to approximate a parameter is called a statistic or an estimator

is the estimator for ̂

statistic parameter

Properties of Good Estimator

Unbiased–

Efficient–

Sufficient–

Consistent–

Estimations & Estimate

Estimation – Is the entire process of using an estimator to produce an estimate of the parameter

2 types of estimation

1. Point Estimate • A single number used to estimate a population parameter

2. Interval Estimate • A spread of values used to estimate a population

parameter• The interval is usually written (a, b) where a and b are

known as confidence limit• a – lower confidence limit• b – upper confidence limit

Definitions

Confidence Interval

– Range of numbers that have a high probability of containing the unknown parameter as an interior point.

– By looking at the width of a confidence interval, we can get a good sense of the estimator precision.

– Width = b – a

Confidence Coefficient

– The probability of correctly including the population parameter being estimated in the interval that is produced

1

Definitions

Level of Confidence

– The confidence coefficient expressed as a percent ,

– Example: 1 100%

1 %

Definitions

OR

(1- α) 100% confidence interval for θ

= ˆ ˆ ˆdistribution for s.d. for

3.3 3.3 Confidence Interval for Mean

OBJECTIVESOBJECTIVES

After completing this chapter, you should be able to

1. Find the confidence interval for the mean.

2. Find the confidence interval for the mean when σ is known and unknown.

Confidence Interval for Mean

Confidence Interval for the Mean

2 2,X z X z

n n

2 2,

s sX z X z

n n

, 1 , 12 2,

n n

s sX t X t

n n

The ( 1 – α ) 100 % confidence interval for μ

t- Distributions

The number of values that are free to vary after a sample statistic has been computed

Rounding Rule

When you are computing a confidence interval for a population mean by using raw data, round off to one more decimal place than the number of decimal places in the original data.

When you are computing a confidence interval for a population mean by using a sample mean and standard deviation, round off to the same number of decimal places as given for the mean.

EXERCISE 3.3

1. The mass of vitamin E in a capsule manufactured by a certain drug company is normally distributed with standard deviation 0.042 mg. A random sample of 5 capsules was analyzed and the mean mass of vitamin E was found to be 5.12 mg. Find the 95% confidence interval for the population mean mass of vitamin E per capsule.

EXERCISE 3.3

2. A plant produces steel sheets whose weights are known to be normally distributed with a standard deviation of 2.4 kg.

A random sample of 36 sheets had a mean weight of 3.14 kg.

Find the 99% confidence interval for the population mean weight.

EXERCISE 3.3

3. A random number of 100 pieces of wood are cut using a machine.

The sample mean of length in cm is 1.06 cm and the standard deviation is 0.08 cm.

a. Find the 90% confidence interval for mean length all the woods cut by the machine.

b. What is the width of this confidence interval?

EXERCISE 3.3

4. The mean IQ score for 25 UMP students is 115 with standard deviation 10.

If the IQ score for all UMP students is normally distributed, find the 95% confidence interval for the mean IQ score for all UMP students.

EXERCISE 3.3

5. The result X of a stress test is known to be normally distributed random variable with mean μ and standard deviation 1.3.

It is required to have a 95% confidence interval for μ with total width less than 2.

Find the least number of tests that should be carried out to achieve this.

EXERCISE 3.3

6. Eight UMP students are randomly chosen and the value of their CPA has been collected as below.

3.20 2.76 2.94 3.41

2.92 2.99 3.01 3.11

Find the 99% confidence interval for the CPA mean for all UMP students.

EXERCISE 3.3

7. The heights of men in a particular district are distributed with mean μ cm and the standard deviation σ cm.

On the basis of the results obtained from a random sample of 100 men from the district, the 95% confidence interval for μ was calculated and found to be (177.22 cm , 179.18 cm). Calculate the value of sample mean and standard deviation.

EXERCISE 3.3

8. A 90% confidence interval for a population mean based on 144 observations is computed to be (2.7, 3.4).

How many observations must be made so that a 90% confidence interval will specify the mean to within ±0.2?

3.4 3.4 Confidence Interval for the Difference Between 2 Mean

OBJECTIVESOBJECTIVES

After completing this chapter, you should be able to

1. Find the confidence interval for the difference between two means when σ’s are known.

2. Find the confidence interval for the difference between two means when σ’s are unknown and equal.

3. Find the confidence interval for the difference between two means when σ’s are unknown and not equal.

Confidence Interval for the Difference Between 2 Mean

EXERCISE 3.4

1. Find the 95% confidence interval for the difference mean of children's sleep time and adults sleep time if given that the variances for children's sleep time is 0.81 hours while for adults is 0.25 hours.

The mean sample sleep time for 30 children's are 10 hours while for 40 adults are 7 hours.

EXERCISE 3.4

2. The mean of sleep time for 50 IPTS students are 7 hours with standard deviation of 1 hour.

The mean of sleep time for 60 IPTA students is 6 hours with standard deviation of 0.7 hour.

Find the 99% confidence interval for the difference mean of sleep time between the IPTS and IPTA students.

a. Assume the population variance are same

b. Assume the population variance are different

EXERCISE 3.4

3. The mean of sleep time for 20 IPTA students are 7.5 hours with variance of 1.11 hours.

The mean of sleep time for 10 IPTS students is 6.3 hours with variance of 0.67 hours.

Find the 95% confidence interval for the difference mean of sleep time between the IPTA and IPTS students.

a. Assume the population variance are same

b. Assume the population variance are not equal

EXERCISE 3.4

4. Two groups of students are given a problem solving test, and the results are compared. The data are as follows:

Mathematics Majors Computer Science majors

Find the 98% confidence interval for the difference mean of test marks between the two groups of students. Assume the variance population test marks are same for both groups.

1

1

1

29

83.6

3.3

n

x

s

2

2

2

28

79.2

2.8

n

x

s

EXERCISE 3.4

5. A medical researcher wishes to see whether the pulse rates of nonsmokers are lower than the pulse rates of smokers. Samples of 110 smokers and 120 nonsmokers are selected. The results are shown here. Smokers Nonsmokers

Find the 90% confidence interval for the difference mean between pulse rates of nonsmokers and the pulse rates of smokers. Assume that the variance pulse rates for both populations are not same.

1

1

1

90

5

110

X

s

n

2

2

2

88

6

120

X

s

n

3.5 3.5 Confidence Interval

for the Proportion OBJECTIVEOBJECTIVE

After completing this chapter, you should be able to

1. Find the confidence interval for a proportion

The ( 1 – α ) 100 % confidence interval for proportion p

2 2,

pq pqp z p z

n n

and 1

ˆ ˆwhere 5 and 5

Xp q p

nnp nq

NOTES:

Sample proportion

Number of successes in a given sample size

Sample size

EXERCISE 3.5

1. 23 from 100 families in a village are poor. Find the 99% confidence interval poorness rate for this village.

2. A survey was undertaken of the use of the internet by residents in a large city and it was discovered that in a random sample of 150 residents, 45 logged on to the internet at least once a day. Calculate an approximate 90% confidence interval for p, the proportion of residents in the city that log on to the internet at least once a day.

3. Given . What sample size is needed to obtain a 95% confidence interval for p with width ± 0.08.

0.3590p

EXERCISE 3.5

4. A researcher whishes to estimate, with 90% confidence, the proportion of people who own a home computer. A previous study shows that 40% of those interviewed had a computer at home. The researcher whishes to be accurate within 5% of the true proportion. Find the minimum sample size necessary.

5. A manufacturer wants to assess the proportion of defective items in a large batch produce by a particular machine. He tests a random sample of 300 items and finds that 45 items are defective. If 200 such tests are performed and a 95% confidence interval calculated for each, what is the probability that more than 194 of the confidence intervals cover the true proportions?

3.6 3.6 Confidence Interval

Difference between Difference between

Two Proportions Two Proportions OBJECTIVEOBJECTIVE

After completing this chapter, you should be able to

1. Find the confidence interval for the difference between two proportions.

The ( 1 – α ) 100 % confidence interval for the difference proportions p1 – p2

1 1 2 21 2

21 2

p q p qp p z

n n

EXERCISE 3.6

1. Given . Find the 95% confidence intervals for .

2. In a sample of 200 surgeons, 15% thought the government should control health care. In a sample of 200 general practitioners, 21% felt the same way. Find 99% confidence interval for the difference of proportions for surgeons and practitioners.

1 1 2 20.6, 75, 0.3, 100p n p n

1 2p p

3.7 3.7 Confidence Interval for Variances and

Standard Deviations

OBJECTIVEOBJECTIVE

After completing this chapter, you should be able to

1. Find the confidence interval for a variance and a standard deviation.

The ( 1 – α ) 100 % confidence interval for the variance

2 2

2 2

, 1 1 , 12 2

1 1,

n n

n s n s

2

2 2

2 2

, 1 1 , 12 2

1 1,

n n

n s n s

The ( 1 – α ) 100 % confidence interval for the standard deviation

Where 22

12

1~ n

n s

(Chi-square distribution)

EXERCISE 3.7

1. A random sample of 10 rulers produce by a machine gives a group of data below, in cm.

100.13, 100.07, 100.02, 99.99, 99.88, 100.14, 100.03, 100.10, 99.92, 100.21

Find the 95% confidence interval for the height variance and standard deviation of all the rulers produce by the machine.

EXERCISE 3.7

2. A factory has a machine that’s designed to filled boxes with an average of 24 ounces of cereal, and the population standard deviation for this filling process is expected to be 0.1 ounce. Thus, if the machine is working properly, the population variance should be 0.01 squared ounce. To estimate the value of population variance, an employee selected a random sample of 15 boxes from a supply filled by the machine and found that the sample variance was 0.008 squared ounce. What’s the 98% confidence interval for the population variance and standard deviation?

3.8 3.8 Confidence Interval for Two Variances and

Standard Deviations

OBJECTIVEOBJECTIVE

After completing this chapter, you should be able to

1. Find the confidence interval for the confidence interval for the variance and a standard deviation proportion.

The ( 1 – α ) 100 % confidence interval

for the variance proportion

Where

2122

2 1

1 2

2 21 12 2 , 1, 122 2, 1, 12

1,

n nn n

s sF

s F s

1 2

2 21 1

1, 12 22 2

~ n n

sF

s

(F distribution)

1 2

2 1

1 , 1, 12, 1, 12

1n n

n n

FF

EXERCISE 3.8

1. The machined in EXERCISE 3.7.1 is serviced. A random sample of 12 rulers produces by the machine after the serviced made give a group of data below.

100.03, 100.01, 100.02, 100.04, 99.90, 99.96, 100.04, 100.06, 100.08, 99.98, 100.11, 100.05

Find the 95% confidence interval for variance proportion for all rulers produces by the machine before and after the service.

EXERCISE 3.8

2. Before service, a machine can packed 10 packets of sugar with variance weight 64 g² while after service the variance weight for 5 packets of sugar are 25 g².

Find the 90% confidence interval for variance proportion for all sugar produces by the machine before and after the service.

Conclusion

An important aspect of inferential statistics is estimation

Estimations of parameters of populations are accomplished by selecting a random sample from that population and choosing and computing a statistic that is the best estimator of the parameter

Statisticians prefer to use the interval estimate rather than point estimate because they can be 95%,99% or else confidence that their estimate contains the true parameter and also determine the minimum sample size necessary.

Thank YouNEXT: CHAPTER 4 HYPOTHESIS TESTING